DEPARTMENT OF MATHEMATICS
181202- MATHEMATICS – II
PART A (Q & A), PART B QUESTION BANK
UNIT-I ORDINARY DIFFERENTIAL EQUATIONS
1.
d2y
dy
 6  13 y  0.
Solve
2
dx
dx
( D 2  6D 13) y  0
A.E is m2 – 6m + 13= 0
6   16
m
 3  2i
2
Hence the solution is y = e3x ( A Cos2x + B Sin2x)
2.
Solve (D2 – 2D + 1) y = ex
A.E is m2 – 2m + 1 =0
(m – 1)2 = 0 => m = 1,1
C.F => (A + Bx) ex
ex
ex
xe x
xe x
x 2e x
P.I = 2




2
D - 2D  1 1  2  1 2D  2 2  2
x 2e x
Solution is y = C.F + P.I = (A + Bx) e +
2
2
Find the particular integral of ( D + 5D + 6) y = 11 e5x
x
3.
P.I =
11e 5 x
D 2  5D  6
Put D = 5 hence P.I =
4
11e 5 x
11e 5 x

25  25  6
56
Solve (D – 1) 2 y = sinh2x.
A.E is (m – 1)2 = 0
m = 1,1
C.F = (A + Bx) ex
sinh 2 x
1
(e 2 x  e 2 x ) 1 e 2 x
1 e 2 x



P.I =
2
2 ( D  1) 2 2 ( D  1) 2
( D  1) 2 ( D  1) 2
=>
1 e2x
1
e 2 x
e 2 x e 2 x



2 (2  1) 2 2 (( 2)  1) 2
2
18
Solution is y = C.F + P.I => (A + Bx) ex +
5.
d2y
dy

4
x
 2y  0
Solve x
dx
dx 2
2
1
e 2 x e 2 x

2
18
6.
7.
(x2 D2 + 4xD + 2) y = 0
Put x = ez ,z = logx, xD =  ,x2 D2 =  ( - 1)
( ( - 1+ 4  + 2) z = 0 => (2 + 3  + 2) z = 0
A.E is m2 + 3m + 2 = 0 => m = -2, -1.
Z = A e-2z + B e-z
A B
=> y = A e-2(logx) + B e-(logx) => y = 2 
x
x
2
Find the particular integral of ( D + 2) y = x2
x2
x2
1
D 2 1 2
P.I = 2


[
1

] x
2
2
D 2
D2
2(1 
)
2
1
D2
= [1 
 ...]x 2
2
2
1 2
P.I = [ x  1]
2
Find the particular integral of (D2 – 2 D + 1)y = coshx
e x  ex
2
x
1 e  ex
x 2e x ex
P.I = [
] 

2 ( D  1) 2
4
8
cosh x 
8.
Solve
dy
dx
 x,
y
dt
dt
Dy – x = 0……….(1)
y – Dx = 0………..(2)
Eliminate x from these two eqns => (D2 – 1)x = 0
A.E is m2 – 1 = 0 => m = 1, -1.
=> x = Aet + B e-t
dx
 Ae t  B e - t = y
=>
dt
Hence x = Aet + B e-t , y = Aet - B e-t
9.
d2y
dy
 4  3 y  sin 3x cos 2 x
Solve
2
dx
dx
(D2 - 4D + 3) y = 0
A.E is m2 – 4m + 3 = 0 => m= 3, m = 1.
C.F = Ae3x + Bex
1
Sin3x cos2x = [sin 5 x  sin x]
2
1
sin 5 x
1
sin x

P.I =
2
2
2 D  4D  3 2 D  4D  3
Put D2 = -52
1
1
[10 cos 5 x  11sin 5 x]  [sin x  2 cos x]
P.I =
884
20
2
1
1
[10 cos 5 x  11sin 5 x]  [sin x  2 cos x]
884
20
2 D2 – (2x + 3) D – 12) y = 6x into linear differential
Transform
the
equation
((2x
+
3)
10.
equation with constant coefficients.
Put 2x + 3 = ez , z = log ( 2x + 3)
( 2x + 3)2 D2 = 4 (2 -  )
( 2x + 3)D = 2
Hence the D.E is (42 - 6 -12) y = z ez - 9
2
3x
11. Find the Particular integral of (D – 3) y = e cosx
cos x
cos x
e 3 x cos x
 e3x
P.I =
= e3x
= -e3x cosx
2
2
2
( D  3  3)
D
( D  3)
2
2
12. Find the Particular integral of (D + 1)y = sin x
y = Ae3x + Bex +
1 - cos2x
2
2
sin x 1 1  cos 2 x 1
1
cos 2 x

 [ 2
 2 ]
P.I = 2
2
2 D 1 D 1
D 1 2 D 1
1 1
P.I =  cos 2 x
2 6
Find
the
Particular integral of (D2 + 4D + 4)y = e-2x x
13.
e 2 x x
x
e 2 x x 3
2 x
2 x 1
P.I =
e
e
( x) =
6
( D  2) 2
(( D  2)  2) 2
D2
sin 2 x 
d2y
dy
2
14. Transform the equation x dx 2  6 x dx  2 y  x log x into linear differential equation
with constant coefficients.
Put x = ez ,z = logx, xD =  ,x2 D2 =  (-1)
the given eqn => ((-1) +6 + 2)y = ez z
(2 + 5 + 2) z = z ez
15. Write Cauchy’s homogeneous linear equation.
dny
d n 1 y
d n2 y
x n n  a1 x n1 n 1  a 2 x n2 n 2  ........  a n y  X
dx
dx
dx
where ai’s are constants and X is a function of x.
16. Write Legendre’s linear equation.
n
n 1
n2
y
y
n d y
n 1 d
n2 d
(a  bx)
 A1 (a  bx)
 A2 (a  bx)
 ........  A n y  f ( x)
n
n 1
n2
dx
dx
dx
A1 , A2,…….are constants.
17. Find the Particular integral of (D + 2) (D – 1)2 = e-2x
1
e 2 x
1 e 2 x 1 e 2 x
xe2 x
[
]


P.I =
=
( D  2) ( D  1) 2
( D  2) 9
9 ( D  2)
9
2
log x 
18. Transform the equation ( x D  xD  1) y  
 into linear differential equation
 x 
2
2
with constant coefficients.
Put x = ez ,z = logx, xD =  ,x2 D2 =  (-1)
3
(2 - 2 + 1) z = (z e-z)2
19. Find the Particular integral of (D2 + 4) y = sin2x
sin 2 x
P.I = 2
D 4
sin 2 x x sin 2 x  x cos 2 x


Put D2 = -22 , P.I =
44
2D
4
2 D2 – 6(3x + 5) D + 8) y = 0 into linear differential
Transform
the
equation
((3x
+
5)
20.
equation with constant coefficients.
Put 3x + 5 = ez , z = log ( 3x + 5)
( 3x + 5)2 D2 = 9 (2 -  )
( 3x + 5)D = 3
Hence the given eqn becomes ( 9 (2 -  ) - 18 + 8 )z = 0 => (92 -27 + 8)z = 0.
PART B
1.
Solve ( D2 + 16 ) y = cos3x
2
Solve by the method of variation of parameters
3.
Solve (x2 D2 – 3 x D + 4 ) y = x2 cos(logx)
dx
dy
 2 y   sin t ,  2 x  cos t given x = 1 and y = 0 at t = 0.
Solve
dt
dt
4.
5.
Solve (x2 D2 – 2 x D - 4 ) y = x2 + 2logx.
6.
Solve
7.
d2y
 4 y  sec 2 x .
dx 2
dx
dy
 2 x  3 y  2e 2t ,  3 x  2 y  0 .
dt
dt
Solve ( D2 + 4D+3 ) y = e–x sinx.
8. Solve ( D2 + 1 ) y = x sinx by the method of variation of parameters.
9. Solve (3x + 2)2 y’’ + 3 (3x + 2) y’ – 36 y = 3x2 + 4x + 1.
10. Solve (D2 -1) y = x ex sinx.
UNIT-II VECTOR CALCULUS
1.
If   3x 2 y  y 3 z 2 , find grad φ at (1, -1, 2)



 6 xy
 3x 2  3 y 2 z 2
 2 y 3 z
x
y
z


grad   6 xy i  3x 2  3 y 2 z 2 j  2 y 3 z k
grad  1, 1, 2   6 i  9 j  4 k
2.
Find  if   2 xz 4  x 2 y at (2, -2, -1).
4
  i



j
k
x
y
z

 
 
 i 2 z 4  2 xy  j  x 2  k 8 xz 3
 2, 2, 1  10 i  4 j  16 k

   100  16  256  372



3.
Find the directional derivative of  = 3x2+2y-3z at (1, 1, 1) in the direction 2i  2 j  k .
4
Find the unit normal vector to the surface x2 + xy + z2 = 4 at the point (1,-1,2).

 
  2i  2 j  k  

  
= 19 .

  n  (6 xi  2 j  3k )  

3

 (1,1,1) 3





 = x2 + xy + z2 ,  (1, 1, 2) = (2 x  y )i  xj  2 zk

 
 i  j  4k
.
 nˆ 


18
5.

(1, 1, 2)



= i  j  4k
Find the angle between the surfaces x logz = y2 – 1 and x2y = 2 –z at the point (1, 1, 1).
Let 1 = y2 –xlogz -1
 

 x
 1 = - log z i  2 yj  k , (  1) (1,1,1) = 2 j  k and |  1| = 5
z
Let 2 = x2y – 2+z



  
 2 = i (2 xy )  j x 2  k (1) , (  2)(1,1,1) = 2i  j  k and |  2| = 6
Cos 
0  2 1
30
   Cos 1  1 
 30 
In what direction from (-1,1,2) is the directional derivative of = xy2z3 a maximum.
Find also the magnitude of this maximum.
=
6.
1.2 ,
1 2
Given = xy2z3






  ( y 2 z 3 )i  (2 xyz 3 ) j  (3xy 2 z 2 )k and  at (1,1,2) = 8i  16 j  12k
derivative occurs in the direction of
 The maximum directional



 = 8i  16 j  12k
The magnitude of this max. directional derivative =  
7.
464

n
n2
Prove that  r  nr r .

 n2


 n1 x
 r n

n2
n
nr
(
x
i

y
j

z
k
) = nr n2 r .
i
nr
x
=  i nr
= 
=
 r  i
r
x





3
3
3
If F  x i  y j  z k , find div curl F
 
 
8.
5

i


curlF    F 
x
x3



j

y
y3

k

z
z3


div (curl F ) =     F = 0
9.



Find ‘a’, such that (3x-2y +z) i  (4 x  ay  z ) j  ( x  y  2 z )k is solenoidal.





Div F = 0  a = –5.

 

10. If A and B are irrotational vectors prove that A  B is solenoidal.




is irrotational  curl B = 0
A isirrotational  curl A = 0 and
B






  ( A  B)  B  ( curl A ) - A  ( curl B ) = B  0 - A  0 = 0.
Div F = (3x-2y +z ) i  (4 x  ay  z ) j  ( x  y  2 z )k = 3 + a + 2 = 5 + a
 
 A  B is solenoidal.



11. Show that the vector F  6 xy  z 3 i  3x 2  z  j  3xz 2  y k is irrotational.



i
j
k




curlF    F 
x
y
z
3
2
6 xy  z 3 x  z 3 xz 2  y
 i (1  1)  j (3 z 2  3 z 2 )  k (6 x  6 x)  0
 F is irrotation al.



12. If F  5 xyi  2 yj , evaluate
 
F
  dr where C is the part of the curve y = x3
C
between x =1 and x=2.
y = x3  dy = 3x2 dx
2
 
4
5
5
6 2
 F  dr =  (5xydx  2 ydy) =  (5x  6 x )dx  ( x  x )1 = 31 + 63 = 94.
1
c
 
 2 c

2
13. If F = x i  xy j , evaluate the line integral  F  dr from (0,0) to (1,1) along the
path y=x.
 
2
2
F
c  dr  c x dx  xy dy =
1
7
.
 x 3 )dx =
12
0





14. If F  (4 xy  3x 2 z 2 )i  2 x 2 j  2 x 3 z k .Check whether the integral F  dr is

 (x
2
C
independent of the path c.
This integral is independent of the path of integration if


F      F  0
6

i


 F 
x
4 xy  3 x 2 z 2

j

y
2x2

k

0
z
 2x3 z
.Hence the line integral is independent of path.
15. State Stoke’s theorem.

If S is an open surface bounded by a simple closed curve C and if a vector function F is

 
continuous and has continuous partial derivatives in S and on C, then  curlF  nˆds =  F  dr
s
c
where n̂ is the unit vector normal to the surface.
16. State Green’s Theorem
If M(x ,y) and N(x ,y) are continuous function with continuous partial derivatives in a region
R of the xy plane bounded by a simple closed curve C, then
 N M 
Mdx

Ndy

C
R  x  y dxdy .Where C is the curve described in positive direction
17. State Gauss Divergence Theorem

If V is the volume bounded by a closed surface S and if a vector function F is continuous
partial derivative in V and on S , then



F nˆ ds   divF dv
S
V
18. Using Divergence theorem, evaluate
 xdydz  ydzdx  zdxdy over the surface of the
s
sphere x2 + y2 + z2 = a2.
By Divergence theorem,

4
=
xdydz

ydzdx

zdxdy


F
dv =  3dv = 3 a 3  = 4a 3 .


3

19. Find the area of a circle of radius ‘a’ using Green’s theorem.
s
v
v
1
xdy  ydx on x2+ y2 =a2 , We have x = a cos  y= asin , :02

c
2
1 2 2
a 2 2
2
2
2
Therefore Area =
(a cos   a sin  )d =
d = a2.


20
2

 0

20. If S is any closed surface enclosing a volume V and F  ax i  byj  czk , prove
 
that  F  nds  (a  b  c)V .
S

 

Div F = a+b+c ;  F  nds   divF  dV = (a+b+c)  dV = (a+b+c)V.
Area =
S
V
V
PART B
1.
Verify stoke’s theorem for F  x i  xy j in the square region in the XY plane bounded by
the lines x = 0, y = 0, x = a and y = a.
2
7
2.
Find the directional derivative of Ф = 2xy + z2 at the point (1,-1,3) in the direction of



i  2 j  2k
3.
Using Gauss – Divergence theorem, evaluate





2
ˆ
A

4
xz
i

y
j

yz
k
where
and
A
.
n
dS

s
S is the surface bounded by the cube x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.
4.
Verify Green’s theorem for
 x
2

dx  xydy where C is the boundary of the square formed by
C
5.
the lines x = 0, y = 0, a = a, y = a.
Show that F  (4xy  z 3 )i  2x 2 j  3xz 2 k is Irrotational and find its scalar potential
6.
Find the work done when a force F  ( x 2  y 2  x )i  (2xy  y) j moves a particle in the xy –
Plane from (0, 0) to (1,1) along the parabola y2 = x.
7.
Find the angle between the normals to the surface xy3z2 =4 at the points (-1,-1,2) and (4,1,-1)
8.
Ir r is the position vector of the point (x,y,z) , Prove that  r  n(n  1)r
Evaluate  ( x 2  xy)dx  ( x 2  y 2 )dy  where C is the square bounded by the line x=0 ,x=1 ,
9.

2
n
n2
C
y =0 and y=1.

10

2
n
n2
Prove that  (r r )  n(n  3)r r
UNIT-III ANALYTIC FUNCTIONS
1.
2.
Define an analytic function (or) holomorphic function (or) Regular function.
A function is said to be analytic at a point if its derivative exists not only at that point but also
in some neighborhood of that point.
Define an entire or an integral function.
3.
A function which is analytic everywhere in the finite plane is called an entire function. An
entire function is analytic everywhere except at z = 
Ex. ez, sinz, cosz, sinhz, coshz
State the necessary condition for f(z) to be analytic [Cauchy – Riemann Equations]
4
5.
The necessary conditions for a complex function f(z) = u(x,y) + iv(x,y) to be analytic in a
region R are
u v
v
u
(i.e) ux = vy and vx = - uy

and

x y
x
y
State the sufficient conditions for f(z) to be analytic
If the partial derivatives ux, uy, vx and vy are all continuous in D and ux = vy and uy= - vx.
Then the function f(z) is analytic in a domain D.
State the polar form of the C – R equation.
In Cartesian co – ordinates any point z is z = x + iy
In polar co – ordinates it is z = rei  where r is the modulus and  is the argument. Then the
C- R equation in polar co – ordinates is given by
8
7.
u 1 v
v
1 u

,

r r 
r
r 
State the basic difference between the limit of a function of a real variable and that of
complex variable.
In the real variable, x →x0 implies that x approaches x0 along the X – axis (or) a line
parallel to the X – axis.
In complex variable, z →z0 implies that z approaches z0 along any path joining the points
z and z0 that lie in the Z – plane.
Define harmonic function.
8.
A real function of two real variables x and y that possesses continuous second order partial
derivatives and that satisfies Laplace equation is called a harmonic function.
Define conjugate harmonic function.
9.
If u and v are harmonic functions such that u + iv is analytic, then each is called the
conjugate harmonic function of the other.
Define conformal transformation.
6.
Consider the transformation w = f(z), where f(z) is a single valued function of z, a point z0
and any two curves C1 and C2 passing through z0 in the Z plane,will be mapped onto a point
w0 and two curves C '1 and C ' 2 in the W plane. If the angle between C1 and C2 at z0 is the
same as the angle between C1' and C 2' at w0 both in magnitude and direction, then the
transformation w = f(z) is said to be conformed at the point z0.
10. Define Isogonal transformation.
A transformation under which angles between every pair of curves through a point are
preserved in magnitude but opposite in direction is said to be isogonal at that point.
11. Define Bilinear transformation.
az  b
The transformation w 
’ ad – bc  0 where a, b,c,d are complex numbers is called a
cz  d
bilinear transformation. This is also called as Mobius or linear fractional transformation.
12. Define Cross Ratio.
z  z 2 z3  z 4  is called the cross
Given four points z1 , z 2 , z3 , z 4 in this order , the ratio 1
z1  z 4 z3  z 2 
ratio of the four points.
2
13. Show that f(z) = z is differentiable at z = 0 but not analytic at z = 0.
Let z = x + iy and z  x  iy
z  zz  x 2  y 2
2
f ( z )  z  ( x 2  y 2 )  i0
2
u  x2  y2 , v  0
u x  2x
, vx  0
uy  2y
, vy  0
So the CR equations ux = vy and uy= - vx are not satisfied everywhere except at z = 0. So f(z)
may be differentiable only at z = 0. Now ux = 2y, vy = 0 and uy = 2y, vx = 0 are continuous
everywhere and in particular at (0, 0). So f(z) is differentiable at z = 0 only and not analytic
there.
9
14. Determine whether the function 2xy + i(x2 – y2) is analytic or not.
Let f(z) = 2xy + i(x2 – y2)
u = 2xy
; v = x2 – y2
ux = 2y, vy = -2y and uy = 2x , vx = 2x
ux  vy and uy  - vx
CR equations are not satisfied.
Hence f(z) is not an analytic function
15. Prove that an analytic function whose real part is constant must itself be a constant.
Let f(z) = u + iv
Given u = constant.  ux = 0 and uy = 0
By CR equations
ux = 0  , vy = 0 ; uy = 0  vx = 0
f1(z) = ux + ivx = 0 + i0
f1(z) = 0  f(z) = c
f(z) is a constant.
16. Show that the function u = 2x – x3 + 3xy2 is harmonic
Given u = 2x – x3 + 3xy2
 2u
 2u
2
2
 6 x ;
ux = 2 – 3x + 3y and
uy = 6xy and
 6x
y 2
x 2
 2u  2u
+
= -6x + 6x = 0
x 2 y 2
Hence u is harmonic
17. Find a function w such that w = u + iv is analytic, if u = exsiny
Given u = exsiny
1 ( x, y )  u x  e x sin y ;  2 ( x, y )  u y  e x cos y
1 ( z,0)  e z (0)  0
;  2 ( z,0)  e z
By Milne Thomson’s method
f ( z )   1 ( z ,0)dz  i   2 ( z ,0)dz  0  i  e z dz  ie z  C
18. Find the image of the hyperbola x2 – y2 = 10 under the transformation w = z2
2
w = z2  u  iv  x  iy   x 2  y 2  i 2 xy
u = x2 – y2 and v = 2xy ; x2 – y2 = 10 (i.e) u = 10
Hence the image of the hyperbola x2 – y2 = 10 under the transformation w = z2 is u = 10
which is a straight line in w plane.
19. Obtain the invariant points of the transformation
2
2
w  2  , The invariant points are given by z  2 
z
z
2z  2
z
 z 2  2z  2  z 2  2z  2  0
z
2  4  8 2  2i
z

 1 i
2
2
20. Define a critical point of the bilinear transformation.
The point at which the mapping w = f(z) is not conformal, (i.e) f1(z) = 0 is called a critical
point of the mapping.
10
PART B
1.
Find the fixed points of the transformation w 
6z  9
z
2
3.
4.
5.
6.
7.
Give an example such that u and v are harmonic but u+iv is not analytic
Prove that ex cos y is harmonic function.
Define bilinear transformation and what the condition for this to be conformal is.
If u + iu is analytic, then prove that v –iu is analytic.
Find the image of the circle |z|=2 under the transformation w=3z.
Determine the region of the w-plane into which the first quadrant of z-plane mapped by
the transformation w=z2.
8. Construct the analytic function f(z) = u+iv given that 2u+3v = ex(cos y – sin y).
9. Find the bilinear transformation that maps z = (1, i, –1) into w=(2, i, –2).
10. Show that ex( x cos y – y sin y) is harmonic function. Find the analytic function f(z) for
which ex (x cos y – y sin y) is the imaginary part.
UNIT-IV COMPLEX INTEGRATION
1.
What is the value of
e
z
dz, where C is Z  1 ?
C
 f ( z )dz  0 if f (z) is analytic inside and on C. Since ez is analytic on and inside
C
the given integral is zero.
2.
3.
cosz
dz where C is Z  2 .
z

1
C
Here f(z) =cos z , C is Z  2 ,
cosz
By Cauchy’s integral Formula 
dz =2 i(cos z ) z 1  2i .
z

1
C
Evaluate
Evaluate


C
e 2z
 z  2 4
dz where C is z  1 using Cauchy’s integral formula.
Z= -2 is out of the given circle z  1 ,
e2z
 z  24 dz  0
C
4.
3z  7 z  1
1
dz, where C is z  .
z 1
2
C
Evaluate 
2
We know that Cauchy’s integral formula is f (a) 
f ( z)
1
dz

2i C z  a
3z 2  7 z  1
3z 2  7 z  1
=
dz
 z  (1) dz, Here f (z) = 3z2+7z+1, a=–1 lies
 z 1
C
C
1
3z 2  7 z  1
outside z  . Therefore by Cauchy’s integral formula 
dz =0.
z 1
2
C
Given
11
Z  1,
5.
Obtain the Laurent expansion of the function
ez
 z  12
in the neighborhood of its
Singular point. Hence find the residue at that point.
Z=1 is a pole of order 2

e 
u2 u3
Put z-1=u .Then f ( z ) 
becomes
f
(
z
)


1

u



...


2! 3!
u2
u2 
z  12

2
3

e. e
e u
u
f ( z)  2   2  
 ...
u u  2! 3!
u

1
Residue of f(z) = coefficient of
=e
u
1
1
Expand f(z)=
in the region Z  1 . Using this result, expand
and tan-1z in
2
1  z 
1 z
e.e u
ez
6.
powers of z.
f ( z )  (1  z) 1 By using binomial series expansions, (1  z )
1
 1  z  z 2  z 3  ...
f ( z )  1  z  z 2  z 3  ...
1
2
3
=1  z  z  z  ... …….(1)
1  z 
1
1

Now
2
1 z
1  ( z 2 )
1
 1  z 2  z 4  z 6  ... -----(2)
Changing z by –z2 in(1)
2
1 z
1
-1
If f(z)=tan z, then f ’(z)=
1 z2
z3 z5 z7
Hence By integrating (2) w.r.to z , tan z= z 


 ...
3
5
7
-1
7.
Obtain the Taylor’s series expansion of log(1+z) when z  0
Let f (z) =log(1+z)
f (0)=log1=0
f ( z ) 
f ( z ) 
1
1 z
1
f ( z ) 
1  z 
f (0) 
1
1
1 0
f (0)  1
2
2
f (0)  2
1  z 
3
12
f iv ( z ) 
6
1  z 
f iv (0)  6
4
f (0)
f (0) 2
z
z  ......
1!
2!
z2 z3 z4
z 2 z3 z 4
log(1  z )  0  z 


 ...... = z     ......
2
3
4
2 3 4
sin z  z
What is the Nature of the singularity at z=0 of the function
.
z3
sin z  z
f(z) =
The function f(z) is not defined at z=0.But by L’Hospital’s rule,
z3
sin z  z
cos z  1
 sin z
 cos z  1
lim z 3  lim 3z 2  lim 6 z  lim 6  6
z 0
z 0
z 0
z 0
log(1  z )  f (0) 
8.
Since the limit exists and is finite, the singularity at z=0 is a removable singularity.
9.
Evaluate

C
Given
z
e 2z
2

1
e 2z
 z 2  1dz
dz where C is z 
1
2
z 2  1 0 : z 2   1
z  i both lies outside c.
c
e 2z
 z 2  1dz  0
Therefore
c
10. Find the residue of the function f(z)=
f ( z) 
4
z ( z  2)
3
4
z 3 ( z  2)
at a simple pole.
, z = 2 is a simple pole. Res[f(z)]z=2 =
1
2
11. State Laurent’s series.
If C1,C2 are two concentric circles with centre at z=a and radii r1 and r2 (r1<r2) and if f(z) is
analytic inside and on the circles and within the annulus between C1 andC2, then for any z in
the annulus, we have f ( z ) 
an 


n 0
n 0
 an ( z  a) n   bn ( z  a) n ,where
f ( z)
f ( z)
1
1
dz
b

dz where C is any circle lying
and
n


2i C ( z  a ) n 1
2i C ( z  a )  n 1
between C1 and C2 with centre at z=a for all n and the integration being taken in positive
direction.
ez
Obtain
the
Laurent
expansion
of
the
function
in the neighborhood of its singular
12.
z2
point. Hence find the residue at that point.
13

ez
1 
z2 z3
becomes
f
(
z
)

1

z


 ...
2
2 
2! 3!
z
z 


1 1 1  z2 z3
f ( z)  2   2  
 ...
z z  2! 3!
z

1
Residue of f(z) = coefficient of
=1
z
1
1
in the region Z  1 Using this result, expand
and
13. Expand f(z)=
1  z 
1 z2
tan-1z in powers of z.
f ( z )  (1  z ) 1 By using binomial series expansions, (1  z ) 1  1  z  z 2  z 3  ...
1
= 1  z  z 2  z 3  ... …….(1)
1  z 
z=0 is a pole of order 2 f ( z ) 
1
1 z
2
 1  z 2  z 4  z 6  ... ----- (2)
If f(z)=tan-1z, then f’(z)=
1
1 z2
z3 z5 z7
Hence By integrating (2) w.r.to z tan z= z 


 ...
3
5
7
-1
14. State Cauchy’s integral formula for nth derivative of an analytic function.
If f(z) is analytic inside and on a simple closed curve C and z=a is any interior point or the
f ( z)
1
dz

2i C  z  a 2
f ( z)
f ( z)
2!
n!
(n)
f (a) 
dz
f
(
a
)

dz ,the integration
and
in
general


2i C  z  a 3
2i C  z  a n 1
region R enclosed by C, then f (a ) 
around C being taken in anti-clockwise direction.
15. Find the region of convergence to expand cos z in Taylor’s series.
Let f(z) = cos z
f
n
z   cos z  n 

2 
 
  n 
f n    cos 

2 
4
4
n 
  n 
f
 
n
cos
 

n



 4 

4
2


f z   
z    
z



n! 
4
n!
4

n 0
n 0
The region of convergence is z    
4
14
16. State Cauchy’s residue theorem.
If f(z) is analytric inside a closed curve C except at a finite number of isolated singular points
a1,a2,…an inside C, then  f ( z )dz = 2i (sum of the residues of f(z) at these singular
C
points).
z
17. Find the poles and residues of f ( z ) 
z 2  3z  2
.
Poles are 1,2
[Res of f(z)]z=1=
lt
( z  1)
lt
( z  2)
z 1
[Res of f(z)]z=2=
z 2
z
 1
( z  1)( z  2)
z
2
( z  1)( z  2)
18. Find the residue of the function f ( z ) 
z2
at a simple pole.
( z  1)( z  2) 2
z2
. Here z = 1 is a simple pole
( z  1)( z  2) 2
[Resf(z)]z=1 = 1
Given f ( z ) 
19. Define essential singularity with an example
If the principal part contains an infinite number of non zero terms, then z=z0 is known as a
essential singularity.
f ( z) 
1
ez
1
1
2
 1  z  z  ... has z=0 as an essential singularity.
1! 2!
Since f(z) is an infinite series of negative powers of z.
20. Define removable singularity with an example.
If the principal part of f(z) contains no terms,That is bn=0 for all n, then the singularity z=z0
is known as the removable singularity of f(z).
Example, f ( z ) 

z2 z4
sin z 1 
z3 z5


 z 

 ... =13
!
5
!
z
z
3! 5!

There is no negative power of z.
Z=0 is a removable singularity.
PART B
z 1
 z 2  2 z  4 dz , where C is z  1  i  2 .
C
1.
Using Cauchy’s integral formula, find
2.
ez
dz , where C is z  3
Using Cauchy’s integral formula, evaluate 
2
(
z

2
)
(
z

1
)
C
15
3.
4
z 2 1
in a Laurent’s series expansion for z  3 and 2  z  3 .
z 2  5z  6
4z
Obtain the Laurent’s series expansion for the function f(z) =
2
 z  1   z  4 in
Expand f(z) =
5.
z 1  4 and 2  z 1  3 .
z2
Find the residues of f(z)=
at its isolated singularities using Laurent’s
( z  1)( z  2) 2
series expansion.
6
Using Cauchy’s residue theorem evaluate
 z
C
7.
Evaluate
 z
e z dz
2
C
2
8
Evaluate
0
Evaluate

2
, where C is z
2
 4

2
, where C is z  i  2
 4 using Cauchy’s residue theorem
d
, using contour integration.
5  4 cos 

2
9.
 2
dz
d
 1  2a cos  a 2  ,
a  1 using contour integration.
0

10. Show that 
0
dx

, using contour integration.

x 1 2 2
4
UNIT- V LAPLACE TRANFORM
1.
State the conditions under which Laplace transform of f(t) exists.
The Laplace transform of f(t) exists if
(i) f(t) is piecewise continuous in the closed interval(a,b) where a >0 .
(ii) f(t) is of exponential order.
2.
Find the Laplace transform of e
1
1
L e 2t t 2  = L t 2 
  ss2


3.
 2t
1
t 2.

  1 
.
=   1
=
3
  2   s  s  2 2( s  2) 2
If L[f(t)] = F(s), prove that L{f(t/5)}= 5 F(5s).
  t   st  t 
L  f   =  e f  dt
5
  5  0
t
put  u  5du = dt
5

  t   5 su
 L  f   =  e
f u 5du = 5  e (5 s )u f u du = 5 F(5s).
  5  0
0
16
4
Find the Laplace transform of unit step function.
0, t  a
The unit step function is defined as u a (t )  
1, t  a, a  0
e  as
and L[ua(t)] =
s
5.
Find the Laplace transform of f(t) = cos2 3t.
L[cos2 3t]
6.
7.
=
L(1)  L(cos 6t )
1
s
s 2  18

=
=
.
2
2s 2( s 2  36)
s( s 2  36)
 cosat 
exist?
 t 
f (t )
cos at 1
lt
 lt
  .
t 0 t
t 0
t
0
 cosat 
does not exist.
 L
 t 
Does L 
Obtain the Laplace transform of sin2t – 2tcos2t in the simplified form.
L[sin 2t  2t cos 2t ] = L[sin 2t ]  2 L[t cos 2t ]
2
d  s 
 (1)  2
= 2

ds  s  4 
s 4
 4 s2 
2

= 2

2 
2

s 4  s 4 
s 2  12
=
2
s2  4
 s2 
Find L1  2
.
 s  2s  2 
 ( s  1)  1 
 s2 
L1  2
= L1 

2

 s  2s  2 
 ( s  1)  1


8.
9.


 ( s  1) 

1
1 
= L1 
L 

2
2
 ( s  1)  1
 ( s  1)  1
  s 
 1 
 L1  2
=e-t  L1  2

 
 s  1 
  s  1
= e-t (cos t + sin t)
What is the Laplace transform of f(t) = f(t +10), 0< t < 10?
Given that f(t) is a periodic function with period 10
1 p st
L{f(t)} =
 e f (t )dt
1  e  ps 0
17
1 10 st
Put p=10, L{f(t)} =
 e f (t )dt
1  e 10 s 0
s2
, find the value of
10. If L{f(t)} = 2
s 4



 f (t )dt =  e
0
 st
0
1 
s

 f (t )dt .
0

1
 s2 
= L[ f (t )]s 0 =  2
=
f (t )dt 

 s  4  s 0 2
 s 0

.
11. Find L 
3
(
s

2
)



 1  s  2  2 
s
 = L 

L1 
3
3 
(s

2
)
(
s

2
)





1
1 
1 
1 


L
= L 
2
2 
 ( s  2) 3 
(
s

2
)




2t 1  1 
 2t 1  2 
= e L  2  – e L  3  = e 2t t (1  t )
s 
s 
12. Find the Laplace transform sin32t
L[sin32t] = ¼ L[3sin2t – sin6t] = ¾ L[sin2t] – ¼ L[sin6t]
=
6 
3 2  1
 2
 -  2

4  s  4  4  s  36 
=
3 1 


2  s2  4 
-
6 1 

 .
4  s 2  36 

 1 
 s 

1 
1  1  
Let F(s) = L  tan   
 s 

1
1
  1
F(s) =
= 2
2  2 
1  (1 / s )  s 
s 1
13. Find L1  tan 1   .
 L1 F ' (s)   sin t ;
L1 F ( s)  


 1 1 '
L F s 
t

 1   sin t
 L1  tan 1    
.
t
 s 

14. Solve using Laplace transform
dy
 y  e t given that y(0)=0.
dt
Taking L.T. on both sides, we get L[y] + L[y] = L[e-t]
18
1
s 1
1
L[y] =
;
( s  1) 2
(s+1) L[y] =

1 
 = t e-t.
2 
 ( s  1) 
1
y = L 
15. Give an example for a function that do not have Laplace transform.
t2
Consider f(t)= e , since lt e
 st t 2
2
e   , hence e t is not of exponential order function.
t 
t2
Hence f(t)= e , does not have Laplace transform.
s3
16. Can F(s)=
s  12
lt F s   lt
s 
s 
s3
be the transform of some f(t)?
s  12
0
Hence F(s) cannot be Laplace transform of f(t).
t
17. Evaluate
 sin u cos(t  u)du .
0


L   sin u cos(t  u )du  Lsin t  cost 
0

t
Let
= L[sint] L[cost]
=
s
s
2
(by convolution theorem)
.

1
2

 t
s
1
=
L
sin
u
cos(
t

u
)
du

  sin t.

2
2
 s  1  2
0
t


18. Give an example for a function having Laplace transform but not satisfying the
continuity condition.
f (t) = t 1 / 2 has Laplace transform even though it does not satisfy the continuity
lt
f (t )  
condition. i.e. It is not piecewise continuous in (0,) as
t 0
19. Define a periodic function and give examples
A function f (t) is said to be periodic function if f(t + p) =f(t) for all t. The least value of p> 0
is called the period of p. For example, sint and cost are periodic functions with period 2
20. State the convolution theorem
t
The convolution of f(t) and g(t) is defined as
 f (u) g (t  u)du and it is denoted by f(t) * g(t).
0
19
PART B
1.
2.
3.
4.
k , 0  t  a
Find the Laplace transform of f(t) = 
and f(t+2a) = f(t) for all t.
 k , a  t  2 a
t , 0  t  a
Find the Laplace transform of f(t) = 
and f(t+2a) = f(t) for all t.
2a  t , a  t  2a
s2
Find the inverse Laplace transform of 2
using convolution theorem.
( s  a 2 )( s 2  b 2 )
Find the Laplace transform of
e
 4t
t
 t sin 3tdt
0
5.
6.
7.
8.
Verify initial and final value theorems for the function f(t) = 1 + e-t(sin t + cos t).
1
Find the inverse Laplace transform of
( s  1)( s 2  4)
Using Laplace transform solve the differential equation y-3y-4y=2e-t with y(0) = y(0) = 1.

Solve the equation y+ 9y=cos2t with y(0) =1 y ( ) = –1.
2

9.
Evaluate  te 2t costdt using Laplace transform.
0
10
Find the Laplace transform of (i) t2 e–tcost (ii) coshat cosat
20